Trevor and his 2 brothers and 5 friends are seated at random in a row of 8 seats at the cinema. What is the probability that Trevor has one brother on his immediate left and one on his immediate right?
i need the workings
To get the special seating configuration, there are 6 possible places for Trevor (he can't be seated in the seats at the edges). Then, you can choose two positions for one given brother of Trevor, the position of the other brother is then fixed.
The remaining 5 friends can then be seated i the remaining 5 seats in 5! ways. So, the total number of ways you can seat Trevor with his two brothers next to him is:
N = 6*2*5!
The total number of ways of seating 8 persons without any conditions is 8!. If you randomly choose one of these 8! configurations, the probability that it will turn out to be one of the N configurations is, of cuorse, N/8!. The probabiblity is thus:
6*2*5!/8! = 6*2*5!/(8*7*6*5!) = 1/(4*7)= 1/28
I looked at it this way.
There are only 2 ways for Trevor (T) and his two friends (F1, F2) to sit beside each other
F1TF2 OR F2TF1
Now consider that triple to be one "item", then we have to arrange 6 "items"
(the 5 friends and the 'triple')
no of ways = 2(6!) = 1440
(the same as Count Iblis' answer)
the no. of ways to arrange without restriction is P(8,6) = 20160
the prob you want is 1440/20160 = 1/14
thanks a lot for the help
where did u get 8*7*6*5 and 6*2*5 from could u jus explain dat
I agree with Count lblis, the denominator should be 8! = 40320.
For the Trevor and his 2 brothers and 5 friends question:
The idea is Trevor cannot sit at the end seats ie seat no. 1 and 8. Imagine sitting at seat 1, you can only have people sitting to your right. (Assuming the seats are arranged like this: s1, s2, s3, s4, s5, s6, s7, s8).
Suppose Trevor sits on seat no. 2. Then of the 8 seats there are only 7 available since Trevor is occupying one by sitting on it. Therefore the first brother will have 1/7 chance of sitting next to him on one side. Now 6 chairs remain. So the other brother will only have 1/6 chance of sitting on the opposite side. Therefore overall: (1/8)*(1/7)*(1/6) = 1/336
Remember: Trevor sat on seat no. 2 but there are 6 seats which he could have sat on (2,3,4,5,6,7) and it would have still worked. So 6*(1/336)= 6/336.
Now one last thing: suppose
Brother 1 = B1
Brother 2= B2
Trevor = T
They can sit like so:
B1, T, B2
This is what we have worked out so far in the above working. The last thing is that they could have sat in the other way i.e.
B2, T, B1
So in total we have to consider 2 different ways of sitting as both would work.
Therefore, 2*(6/336) = 12/336
Simplifying 12/336 = 4/112= 2/56= 1/28
To find the probability that Trevor has one brother on his immediate left and one on his immediate right, we need to determine the total number of favorable outcomes and the total number of possible outcomes.
1. First, let's consider the favorable outcomes. Trevor has 2 brothers, so there are two possible positions for them on his immediate left and immediate right. For each of these positions, we can arrange the remaining 5 friends and 1 additional brother (since Trevor also needs to sit) in the remaining 5 seats. The total number of favorable outcomes is therefore 2 * (5P5 * 1P1) = 2 * 1 = 2.
2. Next, let's consider the total number of possible outcomes. We have a total of 8 people who need to be seated, and we can arrange them in 8P8 = 8! = 40,320 different ways.
3. Finally, we can find the probability by dividing the total number of favorable outcomes by the total number of possible outcomes. Therefore, the probability is 2 / 40,320 ≈ 0.0000496, which can be expressed as approximately 0.005%.
So, the probability that Trevor has one brother on his immediate left and one on his immediate right is approximately 0.005%.